I'm trying to understand the following proof(see picture 1). We need to show that there exists a $y\in X$ such that $\overline{O^+(y)}=X$. But in the whole proof there is no $y$ mentioned . Can anyone explain why the forward orbit of $y$ is dense ?
many thanks in advance
The proof uses the following proposition
Notice that the sequence $\{n_k\}_{k=1}^\infty$ is a sequence of integers (not necessarily positive). Given how the $n_k$ $(k=1,2,\ldots)$ are chosen, we know that $\lim_{k \to \infty} f^{n_k}(x)=x$. Hence $\lim_{k \to \infty} f^l\circ f^{n_k}(x)=f^l(x)$ for any $l \in \mathbb{Z}$. Now for infinitely many positive integers $k$ we must either have that $n_k$ is positive or $n_k$ is negative. If infinitely many $n_k$ are positive then the forward orbit of $x$ is dense in $X$ $\left(\overline{\mathcal{O}^+(x)}=X\right)$. Otherwise, by proposition 2.2.1, the author has shown that $f$ is topologically transitive, and so there is $y \in X$ such that $\overline{\mathcal{O}^+(y)}=X$ (definition of topologically transitive).